As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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**Q1.**We have seen that a gamma-ray does of 3 Gy is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result:

Solution

We can relate an absorbed energy Q and the resulting temperature increase ∆T with relation Q=cm∆T. In that equation, m is the mass of the material absorbing the energy and c is the specific heat of the material. An absorbed dose of 3 Gy corresponds to an absorbed energy per unit mass of 3 J/kg. Let us assume that c the specific heat of human body, is the same as that of water, 4180 J/kg K. Then we find that ∆T=(Q/m)/c=3/4180=7.2×10

We can relate an absorbed energy Q and the resulting temperature increase ∆T with relation Q=cm∆T. In that equation, m is the mass of the material absorbing the energy and c is the specific heat of the material. An absorbed dose of 3 Gy corresponds to an absorbed energy per unit mass of 3 J/kg. Let us assume that c the specific heat of human body, is the same as that of water, 4180 J/kg K. Then we find that ∆T=(Q/m)/c=3/4180=7.2×10

^{-4}K=700Î¼K Obviously the damage done by ionizing radiation has nothing to do with thermal heating. The harmful effects arise because the radiation damages DNA and thus interferes with the normal functioning of tissues in which it is absorbed**Q2.**A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is 36℃. Then the difference of temperature at the two surfaces of A will be

**Q3.**The thermal conductivity of a rod is 2. What is its thermal resistivity??

Solution

Thermal resistivity = 1/(Thermal conductivity) = 1/2 = 0.5

Thermal resistivity = 1/(Thermal conductivity) = 1/2 = 0.5

**Q4.**Newton’s law of cooling is a special case of

Solution

For small difference of temperature, it is the special case of Stefan’s law

For small difference of temperature, it is the special case of Stefan’s law

**Q5.**The spectral energy distribution of a star is maximum at twice temperature as that of sun. the total energy radiated by star is.

Solution

According to Stefan’s law E=ÏƒT

According to Stefan’s law E=ÏƒT

^{4}Where Ïƒ is Stefan’s constant. Given, T=2T_{s}∴E^{'}=Ïƒ(2T_{s})^{4}=16ÏƒT_{s}^{4}=16E_{s}Hence, total energy radiated by star is sixteen times as that of the sun.**Q6.**A particular star (assuming it as a black body) has a surface temperature of about 5×10

^{4}K. The wavelength in nanometers at which its radiation becomes maximum is (b=0.0029 mK)

Solution

According to Wien’s displacement law Î»

According to Wien’s displacement law Î»

_{m}T=b orÎ»_{m}=b/T=0.0029/(5×10^{4})=58×10^{-9}m=58nm**Q7.**On a cold morning, a metal surface will feel colder to touch than a wooden surface because

Solution

Heat passes quickly from the body into the metal which leads to a cold feeling

Heat passes quickly from the body into the metal which leads to a cold feeling

**Q8.**In Searle’s method for finding conductivity of metals, the temperature gradient along the bar

Solution

Is the same at all points along the bar

Is the same at all points along the bar

**Q9.**In which process, the rate of transfer of heat is maximum

Solution

Radiation is the fastest mode of heat transfer

Radiation is the fastest mode of heat transfer

**Q10.**The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is its:

Solution

At boiling point, vapour pressure becomes equal to the external pressure

At boiling point, vapour pressure becomes equal to the external pressure